A) lower the E and shift equilibrium to the left
B) lower the E and shift the equilibrium to the right
C) increase the E and shift the equilibrium to the right
D) increase the E and shift the equilibrium to the left
Correct Answer: C
Solution :
\[Zn(s)\,+2{{H}^{+}}\,\xrightarrow{\,}\,Z{{n}^{2+}}\,(aq)+{{H}_{2}}(g)\] Reaction quotient \[Q=\frac{[Z{{n}^{2+}}]}{{{[{{H}^{+}}]}^{2}}}\] Corresponding cell is \[\underset{Anode}{\mathop{\underset{-}{\mathop{Zn|Z{{n}^{2+}}({{C}_{1}})\,|}}\,}}\,\underset{Cathode}{\mathop{\underset{+}{\mathop{|{{H}^{+}}(aq)|Pt({{H}_{2}})}}\,}}\,\] and \[{{E}_{cell}}\,=E_{cell}^{o}\,-\frac{0.0591}{2}\log K\] \[=E_{cell}^{o}\,-\frac{0.0591}{2}\log \,\frac{[Z{{n}^{2+}}]}{{{[{{H}^{+}}]}^{2}}}\] \[=E_{cell}^{o}\,+\frac{0.0591}{2}\log \,\frac{{{[{{H}^{+}}]}^{2}}}{[Z{{n}^{2+}}]}\] If \[{{H}_{2}}S{{O}_{4}}\] is added to cathodic compartment. (towards reactant side), then Q decreases (due to increase in\[{{H}^{+}}\]). Hence, equilibrium is displaced towards right and \[{{E}_{cell}}\] increases,
You need to login to perform this action.
You will be redirected in
3 sec
