Question
A voltaic cell consists of two half-cells. One half-cell contains a chromium electrode immersed in 1.00 M Cr(NO3)3 solution. The second half-cell contains a nickel electrode immersed in 1.00 M Ni(NO3)2 solution. Nickel plates out on the nickel electrode as the voltaic cell runs. The beginning voltage of the cell is
+0.487 V
at 25°C. The standard electrode potential (standard reduction potential) of chromium at 25°C is
−0.744 V.
(a) Write a balanced half-reaction equation for the reaction occurring at the anode and a second balanced equation for the reaction occurring at the cathode. (Include states-of-matter under the given conditions in your answer.)
anode:
cathode:
(b) Write a balanced equation to show the net reaction for the cell. (Include states-of-matter under the given conditions in your answer.)
(c) Determine the standard electrode potential for nickel.
+0.487 V
at 25°C. The standard electrode potential (standard reduction potential) of chromium at 25°C is
−0.744 V.
(a) Write a balanced half-reaction equation for the reaction occurring at the anode and a second balanced equation for the reaction occurring at the cathode. (Include states-of-matter under the given conditions in your answer.)
anode:
cathode:
(b) Write a balanced equation to show the net reaction for the cell. (Include states-of-matter under the given conditions in your answer.)
(c) Determine the standard electrode potential for nickel.
Answers
You put in the states of matter. The two equations, written as reductions, are below labeled 1 and 2.
1. Cr^3+ + 3e ==> Cr Ered = -0.744 from the problem
2. Ni^2+ + 2e ==> Ni Ered = ? from the problem
Now, if Ni is plating out at the Ni electrode, then obviously rxn 2 must be occurring which means rxn1 is reversed in the cell so here are the eqns as they occur.
3. Cr ==> Cr^3+ + 3e Eox = +0. 744
4. Ni^2+ + 2e ==> Ni Ered = ?
-add 3 and 4 after multiplying to make electrons equal
5, 2Cr + 3Ni^2+ ==> 2Cr^3+ + 3Ni Ecell = Eox + Ered
You know Ecell and Eox. Solve for Ered.
5 gives you part b and part c.
Remember the oxidations occurs at the anode. Look fo reactions 3 and 4 to see which is the anode and which the cathode.
1. Cr^3+ + 3e ==> Cr Ered = -0.744 from the problem
2. Ni^2+ + 2e ==> Ni Ered = ? from the problem
Now, if Ni is plating out at the Ni electrode, then obviously rxn 2 must be occurring which means rxn1 is reversed in the cell so here are the eqns as they occur.
3. Cr ==> Cr^3+ + 3e Eox = +0. 744
4. Ni^2+ + 2e ==> Ni Ered = ?
-add 3 and 4 after multiplying to make electrons equal
5, 2Cr + 3Ni^2+ ==> 2Cr^3+ + 3Ni Ecell = Eox + Ered
You know Ecell and Eox. Solve for Ered.
5 gives you part b and part c.
Remember the oxidations occurs at the anode. Look fo reactions 3 and 4 to see which is the anode and which the cathode.
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