Question

For $Cr_2O_7^{-2} + 14H^{+} + 6e^{-} \to 2Cr^{+3} + 7H_2O; E $$^\circ$= $1.33 V At [Cr_2O_7 ^{-2}] = 4.5$ milli mole $[Cr^{+3}]=15$ milli mole, $E$ is $1.067\, V$.The pH of the solution is nearly equal to

• A. 3
• B. 4
• C. 2
• D. 5

Answer 1

Answer: (C)

$Cr _{2} O _{7}^{2-}+14 H ^{+}+6 e^{-} \longrightarrow 2 Cr ^{3+}+7 H _{2} O$

$E=E^{\circ}-\frac{0.059}{n} \log \frac{\left[ Cr ^{3+}\right]^{2}\left[ H _{2} O \right]^{7}}{\left[ Cr _{2} O _{7}^{2-}\right]\left[ H ^{+}\right]^{14}}$

$1.067=1.33-\frac{0.059}{6}$

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

$\frac{0.059}{6} \log \left[\frac{225 \times 10^{-6}}{\left(4.5 \times 10^{-3}\right) \times\left[ H ^{+}\right]^{14}}\right]=0.263$

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

$=\frac{0.263 \times 6}{0.059}=26.74$

$\log \left[\frac{50 \times 10^{-3}}{\left[ H ^{+}\right]^{14}}\right]=26.74$

$\log \left(50 \times 10^{-3}\right)-\log \left( H ^{+}\right)^{14}=26.74$

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

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

or $+14 pH =28.041$

$pH =\frac{28.041}{14}$

$\left[\because pH =-\log \left[ H ^{+}\right]\right]$

$=2$