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A voltaic cell consists of a Mn/Mn2+ electrode (E° = –1.18 V) and a Zr/Zr4+ electrode...

A voltaic cell consists of a Mn/Mn²⁺ electrode (E° = –1.18 V) and a Zr/Zr⁴⁺ electrode (E° = –1.53 V). Calculate [Mn²⁺] if [Zr⁴⁺ ] = 1.0x10⁻⁴ M and E_cell = 0.40 V at 25°C.

(1) 0.49 M (2) 0.24 M (3) 0.040 M (4) 1.8 M (5) 2.0 M

Expert Solution

Zr(s) -------------------> Zr^4+ (aq) + 4e^- E0 = 1.53v

2Mn^2+(aq) + 4e^- -----------> 2Mn(s) E0 = -1.18v

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

Zr(s) + 2Mn^2+(aq) ----------------> Zr^4+(aq) + 2Mn(s) E0 cell = 0.35v

n = 4

Ecell = E0 cell - 0.0591/n logQ

0.4 = 0.35 - 0.0591/4 log[Zr^4+]/[Mn^2+]^2

0.4-0.35 = -0.0591/4 log[Zr^4+]/[Mn^2+]^2

0.05 = -0.0147 log[Zr^4+]/[Mn^2+]^2

-0.05/0.0147 = log[Zr^4+]/[Mn^2+]^2

-3.4 = log[Zr^4+]/[Mn^2+]^2

[Zr^4+]/[Mn^2+]^2 = 10^-3.4

[Zr^4+]/[Mn^2+]^2 = 0.000398

[Mn^2+]^2 = 1*10^-4/0.000398

[Mn^2+]^2 = 0.25

[Mn^2+] = 0.5M >>>>>>answer

1. 0.49M >>>>answer

[Mn^2+]^2 = 4*10^-8

[Mn^2+] = 2*10^-8 M

queen_honey_blossom answered 1 year ago

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