Question

In: Chemistry

At 25

Expert Solution

+2 0 +2 0

Zn²⁺ + Mn Mn²⁺ + Zn

Given [Mn²⁺] = 0.200 M

[ Zn²⁺] = 0.100 M

E^o = 0.42 V

We know that

$E=E^{o}-\frac{0.059}{n}log\frac{\left [ Mn^{2+}]}{\left [ Zn^{2+}]}$

Where

n = number of electrons transferred = 2

PLug the values we get

cell potential , E = 0.411 V

queen_honey_blossom answered 2 months ago

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