For the cell reaction Zn_{(s)} + 2H^+_{(aq)} | vedclass

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

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$E_{cell}$ increases and the equilibrium shifts to the right.

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(A) The cell reaction is $Zn_{(s)} + 2H^+_{(aq)} 	o Zn^{2+}_{(aq)} + H_{2(g)}$.According to the Nernst equation: $E_{cell} = E^\circ_{cell} - \frac{0.059}{2} \log \frac{[Zn^{2+}] 	imes P_{H_2}}{[H^+]^2}$.This can be rewritten as: $E_{cell} = E^\circ_{cell} + \frac{0.059}{2} \log \frac{[H^+]^2}{[Zn^{2+}] 	imes P_{H_2}}$.Adding $H_2SO_4$ increases the concentration of $H^+$ ions $[H^+]$.As $[H^+]$ increases,the logarithmic term increases,leading to an increase in $E_{cell}$.According to Le Chatelier's principle,an increase in the concentration of reactants $(H^+)$ shifts the equilibrium to the right (towards the products).

For the cell reaction $Zn_{(s)} + 2H^+_{(aq)} \to Zn^{2+}_{(aq)} + H_{2(g)}$,what happens when $H_2SO_4$ is added to the cathode compartment?

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