Oxidation half reaction is

H₂ $$ \to $$ 2H⁺ + 2e^–

If pH = 10, [H⁺] = 10^–10

We know, Nernst Equation

E_cell = E°_cell - $${{0.0591} \over 2}\log {{{{\left[ {{H^ + }} \right]}^2}} \over {{P_{{H_2}}}}}$$

For hydrogen electrode E°_cell = 0

E_cell = - $${{0.0591} \over 2}\log {{{{\left( {{{10}^{ - 10}}} \right)}^2}} \over 1}$$

= 0.0591log10¹⁰

E_cell = 0.591 V

W = $${{IEt} \over {96500}}$$

= $${{10 \times 109 \times 60 \times 59} \over {96500}}$$

= 20

Mn²⁺ + 2e^$$-$$ $$ \to $$ Mn, E^o = $$-$$1.18 V

Mn²⁺ $$ \to $$ Mn³⁺ + e^$$-$$, E^o = $$-$$ 1.51 V

3Mn²⁺ $$ \to $$ Mn^o + 2Mn³⁺, $$\Delta $$E^o = - 1.81 - 1.51 = $$-$$ 2.69 V

Since $$\Delta $$E° is negative,

$$ \therefore $$ $$\Delta $$G = –nFE°, $$\Delta $$G will have positive value so, forward reaction is not possible.

Degree of dissociation($$\alpha $$) = $${{9.54} \over {238}}$$ = 0.04008 = 4.008%