(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_

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

(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.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

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:

AIEEE 2004 All AIEEE

A
Increase the $E_{cell}$ and shift equilibrium to the right
B
Lower the $E_{cell}$ and shift equilibrium to the right
C
Lower the $E_{cell}$ and shift equilibrium to the left
D
Increase the $E_{cell}$ and shift equilibrium to the left

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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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Write a note on the relation between Gibbs free energy and cell potential for a cell reaction.

The cell,$Zn\ |\ Zn^{2+} \,(1\ M)\ ||\ Cu^{2+}\ (1\ M)\ |\ Cu$ $(E^o_{cell} = 1.10\ V)$ was allowed to be completely discharged at $298\ K.$ The relative concentration of $Zn^{2+}$ to $Cu^{2+}$ $\left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$ is

For an electrochemical cell$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

For a galvanic cell reaction at $25^{\circ}C$ with $n = 4$,the standard $emf$ is $0.295 \ V$. What is the equilibrium constant for the reaction? $(F = 96500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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For an electrochemical cell
$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$
the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is
(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)