First slide

Electrochemical cell

Question

The cell reaction of the galvanic cell, Cu_(s) / Cu²⁺_(aq) // Hg²⁺_(aq) / Hg_(l) is

Easy

A

$\large Hg\, + \,C{u^{2 + }}\, \to \,H{g^{2 + }}\, + \,Cu$
B

$Hg\, + \,C{u^{2 + }}\, \to \,H{g^ + }\, + \,C{u^ + }$
C

$Hg\, + \,C{u^ + }\, \to \,CuHg$
D

$Cu\, + \,H{g^{2 + }}\, \to \,C{u^{2 + }} + Hg$

Solution

General cell representation is
Anode // cathode
Negative electrode // Positive electrode
Oxidation half cell //Reduction half cell
Double vertical line represents salt bridge .
At anode Metal looses electrons and undergoes oxidation and at cathode meal ion undergoes reduction.
$\begin{array}{*{20}{c}} {C{u_{\left( s \right)}} \to Cu_{\left( {aq} \right)}^{ + 2} + 2{e^ - }} \\ {Hg_{\left( {aq} \right)}^{ + 2} + 2{e^ - } \to H{g_{\left( s \right)}}} \\ {\overline {\underline {C{u_{\left( s \right)}} + H{g^{ + 2}} \to H{g_{\left( S \right)}} + Cu_{\left( {aq} \right)}^{ + 2}} } } \end{array}$

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