Question

For $Cr _{2} O _{7}^{2-}+14 H ^{+}+6 e ^{-} \rightarrow 2 Cr ^{3+}+7 H _{2} O$
$E ^{0}=+1.33 V$ at $\left[ Cr _{2} O _{7}^{2-}\right]=4.5$ millimole per ltr $\left[ Cr ^{3+}\right]=15$ millimole per ltr , $E$ is $1.06 V$. The $pH$ of the solution is nearly equal to:

• A. 2
• B. 3
• C. 5
• D. 4

Answer 1

Answer: (A)

$E _{\text {cell }}= E _{\text {cell }}^{0}-\frac{0.059}{ n }$ \log $\frac{\left[Cr^{3+}\right]^{2}\left[H_{2}O\right]^{7}}{\left[Cr_{2}O_{7}^{2-}\right]\left[H^{+}\right]^{14}}$

$1.06=1.33-\frac{0.059}{6}$

log $\frac{\left(15 \times 10^{-3}\right)^{2}(1)^{7}}{\left(4.5 \times 10^{-3}\right)\left[ H ^{+}\right]^{14}}$

$0.27=\frac{0.059}{6} \log \left[\frac{225 \times 10^{-6}}{\left(4.5 \times 10^{-3}\right)\left[ H ^{+}\right]^{14}}\right]$

$27.45=\log \left[\frac{50 \times 10^{-3}}{\left[ H ^{+}\right]^{14}}\right]$

$27.45=\log \left[50 \times 10^{-3}\right]-\log \left[ H ^{+}\right]^{14}$

$27.45=-1.3010-14 \log \left[ H ^{+}\right]$

$26.74+1.3010=-14 \log \left[ H ^{+}\right]$

$28.751=14 \times-\log \left[ H ^{+}\right]$

$28.041=14 pH$

$pH =2$