In a cell that utilises the reaction Zn_{(s)} | vedclass

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Question

(A) The cell reaction is $Zn_{(s)} + 2H^{+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + H_{2_{(g)}}$.According to the Nernst equation: $E_{cell} = E^0_

Vedclass · Accepted Answer

Increase the $E_{cell}$ and shift equilibrium to the right

Vedclass · Answer

(A) The cell reaction is $Zn_{(s)} + 2H^{+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + H_{2_{(g)}}$.According to the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[H^{+}]^2}$.When $H_2SO_4$ is added to the cathode compartment,the concentration of $H^{+}$ ions increases.As $[H^{+}]$ increases,the term $\log \frac{[Zn^{2+}]}{[H^{+}]^2}$ decreases,which causes the $E_{cell}$ to increase.According to Le Chatelier's principle,an increase in the concentration of reactants $(H^{+})$ shifts the equilibrium to the right.

In a cell that utilises the reaction $Zn_{(s)} + 2H^{+}_{(aq)} \to Zn^{2+}_{(aq)} + H_{2_{(g)}}$,addition of $H_2SO_4$ to the cathode compartment will:

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