Electrochemistry Question 1
Question 1 - 24 January - Shift 1
At $298 K$, a 1 litre solution containing $10 mmol$ of $Cr_2 O_7{ }^{2-}$ and $100 mmol$ of $Cr^{3+}$ shows a $pH$ of 3.0.
Given : $Cr_2 O_7{ }^{2-} \to Cr^{3+} ; E^{0}=1.330 V$ and
$\frac{2.303 RT}{F}=0.059 V$
The potential for the half cell reaction is $x \times 10^{-3}$ $V$. The value of $x$ is
Show Answer
Answer: (917)
Solution:
$Cr_2 O_7{ }^{2-}+14 H^{+}+6 e^{-} \to 2 Cr^{3+}+7 H_2 O$
$E=1.33-\frac{0.059}{6} \log \frac{(0.1)^{2}}{(10^{-2})(10^{-3})^{14}}$
$E=1.33-\frac{0.059}{6} \times 42=0.917$
$E=917 \times 10^{-3}$
$x=917$
