Question

Plots of $E_{cell}$ with $lo{g}_{10}Q$ for the reaction, $Zn\left(s\right)+C{u}^{2+}(aq.)\rightleftharpoons Cu\left(s\right)+Z{n}^{2+}(aq.)$ gives a straight line with an intercept 1.10. If $E_{cell}=1.1591V$, then value of Q, i.e. $\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$ is:

• A. $0.01$
• B. $0.1$
• C. $100$
• D. $10$

Answer 1

Answer: (A)

$0.01$

According to Nernst equation,

$E=E^o-\frac{0.059}{n}logQ$

From the given reaction,

$Q=\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$ & $n=2$

$E=E^o-\frac{0.059}{2}log\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}=E^o-\frac{0.059}{2}logQ$

$y=c+mx$

intercept $c=E^o=1.10V$

if $E=1.1591V$, then

$1.1591=1.10-\frac{0.059}{2}logQ$

$logQ=\frac{0.059\times 2}{0.059}=-2$

$Q={10}^{-2}$

Option A is correct.