Question

In the cell reaction,
$Cu\left(s\right)+2A{g}^{+}(aq.)\to C{u}^{2+}(aq.)+2Ag\left(s\right)$.
$E_{cell}^{\circ }=0.46V$. By doubling the concentration of $C{u}^{2+},E_{cell}^{\circ }$ is:

• A. Doubled
• B. Halved
• C. Increased but less than double
• D. Decreased by a small fraction

Answer 1

The correct option is D Decreased by a small fraction

Nernst Equation:-

$E_{cell}=E_{cell}^{\circ }-\frac{RT}{nF}lnQ$----------1

$E_{cell}\to$Emf of cell.

$E_{cell}^{\circ }\to$ Standard reduction potential.

$R\to$ Universal gas constant.

$T\to$ Temperature in Kelvin.

$n\to$ moles of electrons.

$F\to$ Faraday's constant.

$Q\to$ Reaction Quotient.

$2A{g}_{\left(aq\right)}^{+}+C{u}_{\left(s\right)}\rightleftharpoons C{u}_{\left(aq\right)}^{2+}+2A{g}_{\left(s\right)}$----------2

$Q=\frac{[Products{]}^a}{[Reactants{]}^b}$

$a$ & $b\to$ stoichiometric co-efficient of product & reactant respectively.

$\therefore \frac{\left(C{u}^{2+}{)}_{aq}\right(A{g}_s{)}^2}{\left(A{g}_{\left(aq\right)}^{+}\right)\left(C{u}_{\left(s\right)}\right)}=\frac{\left[C{u}_{aq}^{2+}\right]}{[A{g}_{aq}^{+}{]}^2}$---------3 [$\frac{[A{g}_s{]}^2}{C{u}_{\left(s\right)}}$=unity in pure form].

$\therefore E_{cell}=E_{cell}^{\circ }-\frac{RT}{nF}ln\frac{\left[C{u}_{aq}^{2+}\right]}{[A{g}_{aq}^{+}{]}^2}$----------4

$\therefore$ From Equation 4, it is observed that $E_{\left(cell\right)}$ depends on the concentration of both $C{u}^{2+}$ & $A{g}^{+}$. It increases with increase in concentration of $A{g}^{+}$.