Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
In a cell\[Zn||Z{{n}^{2+}}||Pt\,{{H}^{+}}|{{H}_{2}},\]the addition of\[{{H}_{2}}S{{O}_{4}}\]to cathode compartment, will (1) decrease the E (2) increase the E (3) shift equilibrium to left (4) shift equilibrium to rightA) 1, 2 and 3 are correct.
B) 1 and 2 are correct.
C) 2 and 4 are correct.
D) 1 and 3 are correct.
Correct Answer: C
Solution :
\[\underset{anode}{\mathop{Zn|Z{{n}^{2+}}}}\,||\underset{cathode}{\mathop{{{H}^{+}}|{{H}_{2}}}}\,(Pt)\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{2}\log \frac{[Z{{n}^{2+}}]}{{{[{{H}^{+}}]}^{2}}}\] \[=E_{cell}^{o}+\frac{0.0591}{2}\log \frac{{{[{{H}^{+}}]}^{2}}}{[Z{{n}^{2+}}]}\] If\[{{H}_{2}}S{{O}_{4}}\]is added to cathodic compartment (towards reactant side), then the reaction shifts in forward direction ie, towards right and\[{{E}_{cell}}\]increases.
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