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← IIT-JEE 2010 · Paper 1
ChemistrySingle correct+3 / -1Question 18 of 84

IIT-JEE 2010 · Paper 1, Chemistry question 18

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Paragraph for Questions 17 to 18

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is:

M(s) ∣ MX+(aq; 0.05 molar) ∣∣ MX+(aq; 1 molar) ∣ M(s)\ce{M(s) | M+(aq; 0.05 molar) || M+(aq; 1 molar) | M(s)}

For the above electrolytic cell the magnitude of the cell potential ∣Ecell∣=70|E_{cell}|=70 mV.

If the 0.05 molar solution of MX+\ce{M+} is replaced by a 0.0025 molar MX+\ce{M+} solution, then the magnitude of the cell potential would be

Choose one option
Answer and worked solution

Official answer: C

Nernst equation. E=0.059log⁡1[MX+]anodeE=0.059\log\dfrac{1}{[\ce{M+}]_\text{anode}}.

With 0.05 M the ratio is 20; with 0.0025 M it is 400=202400=20^2. The logarithm doubles, so the potential doubles:
∣E∣=2×70=140 mV.|E|=2\times70=140\text{ mV}.

(C)