COMEDK · Chemistry · 18. Electrochemistry
For a cell \(2 \mathrm{M}_{(\mathrm{S})}+\mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}\) the \(\mathrm{E}^0\) cell \(=+1.67 \mathrm{~V}\). When \(\left[\mathrm{M}^{2+}\right]\) is \(1.0 \times 10^{-3} \mathrm{M}\) and \(\mathrm{p}\left(\mathrm{O}_2\right)\) is 0.1 atm , the EMF of the cell becomes +1.57 V . Calculate the pH of the electrochemical cell.
- A \(3.49\)
- B \(12.01\)
- C \(5.24\)
- D \(2.94\)
Answer & Solution
Correct Answer
(D) \(2.94\)
Step-by-step Solution
Detailed explanation
The cell reaction is \(2M_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2M^{2+}_{(aq)} + 2H_2O_{(l)}\). The Nernst equation for the cell is \(E_{cell} = E^0_{cell} - \dfrac{0.0591}{n} \log Q\), where \(n=4\) electrons are transferred. The reaction quotient \(Q\) is given by…
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