Question

Q. there are two possible reaction for cathode in the electrolysis of aqueous ZnCl2:
Zn2 + (aq) + 2e- gives Zn(s). E=-0.76v
2H2O(1) + 2e- gives H2(g) + 2 OH-(aq). E=-0.83v
which one will take place?​

Answer 1

Answer:

In the electrolysis of aqueous ZnCl₂, the following occurs -
$Zn^2^+ _(_a_q_) + 2e^- \rightarrow Zn_(_s_) \;\;\;\;(E=-0.76V)$

Explanation:

ZnCl₂ doesn't conduct electricity in the solid-state, but can conduct electricity in its molten state.

At the cathode, the following reactions could possibly occur -

$Zn^2^+ _(_a_q_) + 2e^- \rightarrow Zn_(_s_) \;\;\;(E=-0.76V)\\2H_2O_(_l_) + 2e^- \rightarrow H_2 _(_g_) + 2 OH^- _(_a_q_) \;\;\; (E=-0.83V)$

However, out of the two, the first reaction will occur, because it has a greater value of E°.

(We know that a greater the value of E° indicates the greater possibility of reduction occuring for the reaction at the cathode.)
Thus, the reduction of Zn²⁺ will take place at cathode to form Zn metal which gets deposited on the cathode.

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