Electrochemistry 2 Question 29
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28. If the 0.05 molar solution of $M^{+}$is replaced by a 0.0025 molar $M^{+}$solution, then the magnitude of the cell potential would be
======= ####28. If the 0.05 molar solution of $M^{+}$is replaced by a 0.0025 molar $M^{+}$solution, then the magnitude of the cell potential would be
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $35 \mathrm{mV}$
(b) $70 \mathrm{mV}$
(c) $140 \mathrm{mV}$
(d) $700 \mathrm{mV}$
Passage III
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential $\left(E^{\circ}\right)$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}(V$ with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 17-19.
$(2007,4 \times 3 \mathrm{M}=12 \mathrm{M})$
$$ \begin{aligned} \mathrm{I}{2}+2 e^{-} \rightarrow 2 \mathrm{I}^{-} & E^{\circ}=0.54 \ \mathrm{Cl}{2}+2 e^{-} \rightarrow 2 \mathrm{Cl}^{-} & E^{\circ}=1.36 \ \mathrm{Mn}^{3+}+e^{-} \rightarrow \mathrm{Mn}^{2+} & E^{\circ}=1.50 \ \mathrm{Fe}^{3+}+e^{-} \rightarrow \mathrm{Fe}^{2+} & E^{\circ}=0.77 \ \mathrm{O}{2}+4 \mathrm{H}^{+}+4 e^{-} \rightarrow 2 \mathrm{H}{2} \mathrm{O} & E^{\circ}=1.23 \end{aligned} $$
Show Answer
Solution:
- $E_{\text {cell }}=E^{\circ}-\frac{0.0538}{1} \log 0.0025=0.139 \mathrm{~V} \approx 140 \mathrm{mV}$
