Calculate the cell potential for a reaction in a electrolytic cell with the following half-reactions if:...

Calculate the cell potential for a reaction in a electrolytic cell with the following half-reactions if: [U 3+] = 0.10 M, [MnO4 - ] = 0.20M, [Mn2+], and [H+ ] = 0.20 M

U 3+ + 3e -> U o Ecell = - 1.642 V

MnO4 - + 8H+ + 5e- -> Mn2+ + 4H2O Ecell = +1.51 V

Solutions

Expert Solution

U 3+ + 3e -> U o Ecell = - 1.642 V

MnO4 - + 8H+ + 5e- -> Mn2+ + 4H2O Ecell = +1.51 V

5U-----------> 5U^3+ (aq) + 15e^- Ecell = 1.642v

3MnO4 - + 24H+ + 15e- -> 3Mn2+ + 12H2O Ecell = +1.51 V

-----------------------------------------------------------------------------------------------------------------

5U + 3MnO4^- + 24H^+ -------------> 5U^3+ (aq) + 3Mn^2+ + 12H2O Ecell = 3.152V

n = 15

Ecell = E0cell - 0.0592/n logqQ

= 3.152 - 0.0592/15 log[U^3+]^5 [Mn^2+]^3/[MnO4^-]^3 [H^+]^24

= 3.152 -0.00394 log(0.1)^5(0.2)^3/(0.2)^3(0.2)^24

= 3.152 - 0.00394log(596046447753)

= 3.152-0.00394*11.7752

= 3.1056V >>>>answer

queen_honey_blossom answered 1 year ago

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