JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
One half cell in a voltaic cell is constructed by dipping silver rod in \(AgNO_3\) solution of unknown concentration, other half cell is \(Zn\) rod dipped in \(1\) molar solution of \(ZnSO_4\). A voltage of \(1.60\text{ V}\) is measured at \(298\text{ K}\) for this cell. What is the concentration of \(Ag^+\) ions used in terms of \(\log x\) (\(x = [Ag^+]\)) ?
\(E^\circ_{Zn^{2+}/Zn} = -0.76\text{ V}\), \(E^\circ_{Ag^+/Ag} = +0.80\text{ V}\), \(\dfrac{2.303 RT}{F} = 0.059\text{ V}\)
- A \(\dfrac{2}{3.9}\)
- B \(\dfrac{4}{5.9}\)
- C \(\dfrac{2.9}{2}\)
- D \(\dfrac{5.9}{4}\)
Answer & Solution
Correct Answer
(B) \(\dfrac{4}{5.9}\)
Step-by-step Solution
Detailed explanation
The cell reaction is \(Zn(s) + 2Ag^+(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)\) The standard cell potential is given by: \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-0.76) = 1.56\text{ V}\) Using the Nernst equation:…
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