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.
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.
1 answer
1 answer