Question

A voltaic cell contains two half-cells. One half-cell contains a chromium electrode immersed in a 1.00 M Cr(NO_3)_3 solution. The second half-cell contains a nickel electrode immersed in a 1.00 M Ni(NO_3)_2 solution. Cr^3+(aq) + 3 e^- rightarrow Cr(s) E_red^degree = 0.744 V Ni^2+(aq) + 2e^- rightarrow Ni(s) E_red^degree = -0.257 V Using the standard reduction potentials given above, predict the standard cell potential of the voltaic cell. Write the overall balanced equation for the voltaic cell. (Include states-of-matter under the given conditions in your answer.)

Answer 1

a)

E°Cell = Ecathode - Eanode

cathode --> Nickel, since has higher potential

anode --> Chrmomium since has lower potential

so..

E°Cell = (-0.257) - (-0.744) = 0.487 V

b)

the balanced cell:

Cr+3 + 3e- --> Cr(s)

Ni2+ + 2e- --> Ni(s)

invert the one being oxidized, i.e. Chromium

Ni2+ + 2e- --> Ni(s)

Cr(s) --> Cr+3 + 3e-

balance e-

3Ni2+ + 6e- --> 3Ni(s)

2Cr(s) --> 2Cr+3 + 6e-

add all

3Ni2+ + 6e- + 2Cr(s)--> 3Ni(s) + 2Cr+3 + 6e-

cancle common terms

3Ni2+ + 2Cr(s)--> 3Ni(s) + 2Cr+3