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.
Part B
A) The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell is the site of the reduction reaction.
Type the half-cell reaction that takes place at the anode for the chromium-silver voltaic cell. Indicate the physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not forget to add electrons in your reaction.
Express your answer as a chemical equation.
B) The half-cell is a chamber in the voltaic cell where one half-cell is the site of an oxidation reaction and the other half-cell is the site of a reduction reaction.
Type the half-cell reaction that takes place at the cathode for the chromium-silver voltaic cell. Indicate physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not forget to add electrons in your reaction.
To determine the half-cell reactions that take place at the anode and cathode for the chromium-silver voltaic cell, we need to consider the oxidation and reduction reactions occurring at each electrode.
A) The anode is the site of the oxidation reaction. In this case, the chromium electrode (Cr) is oxidized. The Cr(NO3)3 solution supplies the necessary ions for the oxidation to occur. The half-cell reaction at the anode is:
Cr(s) -> Cr^3+(aq) + 3e-
B) The cathode is the site of the reduction reaction. In this case, the silver electrode (Ag) is reduced. The AgNO3 solution supplies the necessary Ag+ ions for reduction to occur. The half-cell reaction at the cathode is:
Ag+(aq) + e- -> Ag(s)
To determine the half-cell reactions that occur at the anode and cathode in the chromium-silver voltaic cell, we need to consider the oxidation and reduction reactions.
First, let's determine the half-cell reaction that takes place at the anode (site of the oxidation reaction).
Step 1: Identify the species involved at the anode. In this case, the anode consists of the chromium (Cr) electrode immersed in a Cr(NO3)3 solution.
Step 2: Write the balanced chemical equation for the oxidation of the species at the anode. The oxidation of chromium (Cr) can occur in an acidic solution (Cr(NO3)3). The balanced half-cell reaction is as follows:
Cr(s) → Cr3+(aq) + 3e-
In this reaction, solid chromium (Cr) is oxidized to chromium ions (Cr3+) with the release of 3 electrons (e-).
Next, let's determine the half-cell reaction that takes place at the cathode (site of the reduction reaction).
Step 1: Identify the species involved at the cathode. In this case, the cathode consists of the silver (Ag) electrode immersed in an AgNO3 solution.
Step 2: Write the balanced chemical equation for the reduction of the species at the cathode. The reduction of silver ions (Ag+) can occur in an aqueous solution (AgNO3). The balanced half-cell reaction is as follows:
Ag+(aq) + e- → Ag(s)
In this reaction, silver ions (Ag+) are reduced and gain one electron (e-) to form solid silver (Ag).
Therefore, the half-cell reactions for the chromium-silver voltaic cell are as follows:
Anode (Oxidation): Cr(s) → Cr3+(aq) + 3e-
Cathode (Reduction): Ag+(aq) + e- → Ag(s)
I see only parts A and B, no C nor D.
A is the oxidation at the anode of Cr.
Cr(s) ==> Cr^3+(aq) + 3e
B is the reduction at the cathode.
3Ag^+(aq) + 3e==> 3Ag(s)