1) Assuming the following reaction proceeds in the forward direction,

3 Sn4+(aq) + 2 Cr(s) ---> 3 Sn2+(aq) + 2 Cr3+(aq)
a.Sn4+(aq) is the reducing agent and Cr(s) is the oxidizing agent.

b.Cr(s) is the reducing agent and Sn2+(aq) is the oxidizing agent.

c.Sn4+(aq) is the reducing agent and Sn2+(aq) is the oxidizing agent.

d.Cr(s) is the reducing agent and Cr3+(aq) is the oxidizing agent.

e.Cr(s) is the reducing agent and Sn4+(aq) is the oxidizing agent.

The answer C

2) Write a balanced chemical equation for the oxidation of Cd(s) by concentrated nitric acid, producing NO2(g) and Cd2+(aq).
a. HNO3(aq) + Cd(s) ---> Cd2+(aq) + NO2(g) + OH–(aq)
b. 2 HNO3(aq) + Cd(s) ---> Cd2+(aq) + 2 NO2(g) + 2 OH–(aq)
c. HNO3(aq) + Cd(s) + H+(aq) ---> Cd2+(aq) + NO2(g) + H2O()
d. 4 HNO3(aq) + Cd(s) ---> Cd2+(aq) + 2 NO2(g) + 2H2O() + 2 NO3–(aq)
e. HNO3(aq) + Cd(s) ---> Cd2+(aq) + NO2(g)

the answer is D

3) Consider the following half-reactions:
Cl2(g) + 2 e– -> 2 Cl–(aq) E = +1.36 V
Ag+(aq) + e– -> Ag(s) E = +0.80 V
Cu2+(aq) + 2 e– -> Cu(s) E = +0.34 V
Sn2+(aq) + 2 e– -> Sn(s) E = –0.14 V
Al3+(aq) + 3 e– -> Al(s) E = –1.66 V
Which of the above elements or ions will reduce Cu2+(aq)?
a. Ag(s) and Sn2+(aq)
b. Cl–(aq) and Ag(s)
c. Cl2(g) and Ag+(aq)
d. Sn(s) and Al(s)
e. Sn2+(aq) and Al3+(aq)

the answer is D

4)Given the following two half-reactions, write the overall balanced reaction in the direction in which it is spontaneous and calculate the standard cell potential.
Ga3+(aq) + 3 e– -> Ga(s) E = –0.53 V
Sn4+(aq) + 2 e– -> Sn2+(aq) E = +0.15 V
a. 2 Ga3+(aq) + 3 Sn2+(aq) -> 2 Ga(s) + 3 Sn4+(aq) = +0.68 V
b. 3 Ga3+(aq) + 2 Sn2+(aq) ->3 Ga(s) + 2 Sn4+(aq) = –1.89 V
c. 2 Ga(s) + 3 Sn4+(aq) -> 2 Ga3+(aq) + 3 Sn2+(aq) = +0.68 V
d. 3 Ga(s) + 2 Sn4+(aq) -> 3 Ga3+(aq) + 2 Sn2+(aq) = +0.38 V
e. 2 Ga(s) + 3 Sn4+(aq) -> 2 Ga3+(aq) + 3 Sn2+(aq) = +1.89 V

the answer is D

1. Cris, I don't agree with C as an answer. Here is the way you keep these straight. Just remember the definition of oxidation and reduction.

Oxidation is the loss of electrons.
Reduction is the gain of electrons.
The substance oxidized is the reducing agent.
The substance reduced is the oxidizing agent.

2 is ok

3 is ok
I don't agree with 4.

To determine the reducing and oxidizing agents in a given reaction, we need to look at the changes in oxidation states of the elements involved.

In the reaction:

3 Sn4+(aq) + 2 Cr(s) ---> 3 Sn2+(aq) + 2 Cr3+(aq)

The Sn4+ ion is going from an oxidation state of +4 to +2, which means it is undergoing reduction and is therefore the oxidizing agent.
The Cr atom is going from an oxidation state of 0 to +3, which means it is undergoing oxidation and is therefore the reducing agent.

So, the correct answer is:

c. Sn4+(aq) is the reducing agent and Sn2+(aq) is the oxidizing agent.

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To write a balanced chemical equation for the oxidation of Cd(s) by concentrated nitric acid, producing NO2(g) and Cd2+(aq), we need to determine the stoichiometric coefficients of the reactants and products.

The correct balanced chemical equation is:

4 HNO3(aq) + Cd(s) ---> Cd(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)

So, the correct answer is:

d. 4 HNO3(aq) + Cd(s) ---> Cd2+(aq) + 2 NO2(g) + 2H2O(l) + 2 NO3–(aq)

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To determine which elements or ions will reduce Cu2+(aq), we need to compare their standard reduction potentials (E°) with that of Cu2+.

The reduction potential of Cu2+/Cu is +0.34 V.
From the given options, the only element or ion with a reduction potential higher than 0.34 V is Sn(s) (-0.14 V). Therefore, Sn(s) will reduce Cu2+(aq).

So, the correct answer is:

d. Sn(s) and Al(s)

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To write the overall balanced reaction and calculate the standard cell potential, we need to consider the reduction potentials (E°) of the half-reactions. The overall reaction will be spontaneous in the direction where the cell potential is positive.

The given half-reactions are:

Ga3+(aq) + 3 e– -> Ga(s) E° = –0.53 V
Sn4+(aq) + 2 e– -> Sn2+(aq) E° = +0.15 V

To obtain a positive cell potential, the half-reaction with the higher reduction potential (Sn) needs to be reversed and multiplied by the appropriate coefficient. Then, the two half-reactions can be added together.

The correct overall balanced reaction is:

3 Ga(s) + 2 Sn4+(aq) -> 3 Ga3+(aq) + 2 Sn2+(aq)

To calculate the standard cell potential, we sum the reduction potentials of the half-reactions:

E°cell = E°reduction (Sn2+/Sn) - E°oxidation (Ga3+/Ga)
= +0.15 V - (-0.53 V)
= +0.68 V

So, the correct answer is:

a. 2 Ga3+(aq) + 3 Sn2+(aq) -> 2 Ga(s) + 3 Sn4+(aq) = +0.68 V

To answer these questions, we need to understand the concepts of oxidation and reduction, as well as how to determine which species acts as the reducing agent and which acts as the oxidizing agent.

1) In this reaction, Sn4+(aq) is being reduced to Sn2+(aq) and Cr(s) is being oxidized to Cr3+(aq). The reducing agent is the species that is being oxidized, while the oxidizing agent is the species that is being reduced. Therefore, the correct answer is c. Sn4+(aq) is the reducing agent and Sn2+(aq) is the oxidizing agent.

2) To write a balanced chemical equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. From the reaction, we can see that HNO3(aq) is being reduced to NO2(g) and Cd(s) is being oxidized to Cd2+(aq). The balanced equation is 4 HNO3(aq) + Cd(s) -> Cd2+(aq) + 2 NO2(g) + 2 H2O(). So, the correct answer is d.

3) To determine which species will reduce Cu2+(aq), we need to compare the standard reduction potentials (E°) of Cu2+(aq) with the other elements or ions. The species with a higher reduction potential will be able to reduce Cu2+(aq). In this case, only Sn(s) and Al(s) have higher reduction potentials than Cu2+(aq). So, the correct answer is d. Sn(s) and Al(s) will reduce Cu2+(aq).

4) To write the overall balanced reaction and calculate the standard cell potential, we need to combine the two given half-reactions. It is important to remember that the electrons need to cancel out in the overall reaction. From the given half-reactions, we can see that Ga3+(aq) is being reduced to Ga(s) and Sn4+(aq) is being reduced to Sn2+(aq). By combining these half-reactions, we get: 3 Ga(s) + 2 Sn4+(aq) -> 3 Ga3+(aq) + 2 Sn2+(aq). The standard cell potential is calculated by adding the reduction potentials of the half-reactions. In this case, the reduction potentials for Ga3+(aq) and Sn4+(aq) are -0.53 V and +0.15 V, respectively. So, the correct answer is d. 3 Ga(s) + 2 Sn4+(aq) -> 3 Ga3+(aq) + 2 Sn2+(aq) with a standard cell potential of +0.38 V.