Consider a chromium-silver voltaic cell that is constructed such that one half-cell consists of the chromium, Cr, electrode immersed in a Cr(NO3)3 solution, and the other half-cell consists of the silver, Ag, electrode immersed in a AgNO3 solution. The two electrodes are connected by a copper wire. The Cr electrode acts as the anode, and the Ag electrode acts as the cathode. To maintain electric neutrality, you add a KNO3 salt bridge separating the two half-cells. Use this information to solve Parts B, C, and D.What is the net cell reaction for the chromium-silver voltaic cell?
Were are parts B, C, and D.
Rxn is (add the two half cells below)
Cr(s) ==> Cr^3+ + 3e
Ag^+ + e ==> Ag(s)
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Cr(s) + 3Ag^+ ==> 3Ag(s) + Cr^3+
Well, I have to say, this is quite shocking! But don't worry, I've got you covered. The net cell reaction for the chromium-silver voltaic cell can be determined by writing the half-cell reactions and balancing the electrons.
The half-cell reactions can be written as follows:
At the anode (Cr electrode): Cr(s) → Cr3+(aq) + 3e-
And at the cathode (Ag electrode): Ag+(aq) + e- → Ag(s)
To balance the electrons, we need to multiply the cathode reaction by 3:
3Ag+(aq) + 3e- → 3Ag(s)
Now, let's combine the anode and cathode reactions to get the net cell reaction:
Cr(s) + 3Ag+(aq) → Cr3+(aq) + 3Ag(s)
Voilà! That's the net cell reaction for the chromium-silver voltaic cell. Keep in mind that the silver ions play the role of oxidizing agent (being reduced) while the chromium acts as the reducing agent (being oxidized).
To determine the net cell reaction for the chromium-silver voltaic cell, we need to write the half-cell reactions and then balance them. The half-cell reactions are as follows:
At the chromium (Cr) electrode (anode):
Cr(s) -> Cr^3+(aq) + 3e-
At the silver (Ag) electrode (cathode):
Ag+(aq) + e- -> Ag(s)
To balance these half-cell reactions, we need to multiply them by appropriate coefficients so that the number of electrons gained in the cathode reaction matches the number of electrons lost in the anode reaction.
Multiplying the cathode reaction by 3 and the anode reaction by 1, we get:
Cr(s) -> Cr^3+(aq) + 3e-
3Ag+(aq) + 3e- -> 3Ag(s)
Now, we can add these two reactions together to obtain the overall cell reaction:
Cr(s) + 3Ag+(aq) -> Cr^3+(aq) + 3Ag(s)
Thus, the net cell reaction for the chromium-silver voltaic cell is:
Cr(s) + 3Ag+(aq) -> Cr^3+(aq) + 3Ag(s)
To determine the net cell reaction for the chromium-silver voltaic cell, we need to consider the oxidation and reduction half-reactions that occur at the anode and cathode, respectively.
First, let's identify the half-reaction at the anode. In this case, the chromium electrode is immersed in a Cr(NO3)3 solution and acts as the anode. Chromium is being oxidized, so its oxidation half-reaction can be written as follows:
Cr(s) → Cr3+(aq) + 3e-
Next, let's identify the half-reaction at the cathode. The silver electrode is immersed in an AgNO3 solution and acts as the cathode. Silver is being reduced, so its reduction half-reaction can be written as follows:
Ag+(aq) + e- → Ag(s)
Now, we need to balance the half-reactions to ensure that the number of electrons transferred is the same in both half-reactions. To do this, multiply the oxidation half-reaction by 1 and the reduction half-reaction by 3:
Cr(s) → Cr3+(aq) + 3e-
3Ag+(aq) + 3e- → 3Ag(s)
Next, add the two half-reactions together to obtain the net cell reaction:
Cr(s) + 3Ag+(aq) → Cr3+(aq) + 3Ag(s)
So, the net cell reaction for the chromium-silver voltaic cell is:
Cr(s) + 3Ag+(aq) → Cr3+(aq) + 3Ag(s)