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The reduction potential of hydrogen half-cell will be negative if

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a

p(H2)=1 atm and H+=2.0 M

b

p(H2)=1 atm and H+=1.0 M

c

p(H2)=2 atm and H+=1.0 M

d

p(H2)=2 atm and H+=2.0 M

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detailed solution

Correct option is C

Reduction hydrogen half-cell is H+zMPt(H2);Pressure pH2Half-cell reaction is 2H+aq+2e-→H2(g)Reaction quotient =Q=pH2/H+2and n=2Ecello=Eredo-0.0591nlog Q=0-0.05912log QEredo=0.00 Vfor standard hydrogen electrode If Q>1,then Eredo=-ve Thus, correct answer is (c ).

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Similar Questions

The reaction which occur in the Galvanic cell is
$\begin{matrix} {MnO}_{4}^{-} + 8 H^{+} + 5 {Fe}^{2 +} ⟶ {Mn}^{2 +} + 5 {Fe}^{3 +} + 4 H_{2} O \\ E^{\circ} ({MnO}_{4}^{-}, {Mn}^{2 +}, H^{+} ∣ Pt) = 1 .51 V and \\ E^{\circ} ({Fe}^{3 +}, {Fe}^{2 +} ∣ Pt) = 0 .77 V \end{matrix}$
(Q). If $[H_{3} O^{+}] = 2 .0 M, [{Fe}^{3 +}] = 2 .0 M, [{Fe}^{2 +}] = 0 .5 M$ $[{Mn}^{2 +}] = 0 .5 M, [{MnO}_{4}^{-}] = 1 .0 M$ Calculate the emf of the cell.

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