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

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.

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

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

= 20

At anode :Zn²⁺_(aq) + 2e^$$-$$ $$ \to $$ Zn_(s);  E^o = $$-$$0.76 V

At cathode : Ag₂O_(s) + H₂O_(l) + 2e^$$-$$ $$ \to $$ 2Ag_(s) + 2OH^$$-$$_(aq), E^o = 0.34 V

E°_cell = E°_cathode – E°_anode

= 0.34 – (–0.76) = 1.10 V