Question

Observe the following cell \[ M(s)\;|\;M^{2+}(xM)\;||\;H^+(0.02M)\;|\;H_2(g,1\,bar)\;,\;Pt(s) \] What is the value of \(x\)? Given: \[ \frac{2.303RT}{F}=0.06V \] \[ E^\circ_{M^{2+}|M}=-0.14V \] \[ E^\circ_{H^+|H_2}=0.0V \] \[ E_{cell}=0.077V \] \[ \log 4=0.602 \] \[ \text{antilog}(2.7)=0.05,\qquad \text{antilog}(2.60)=0.04 \]

• A. \(0.05\)
• B. \(0.04\)
• C. \(0.002\)
• D. \(0.001\)

Hint:

For electrochemical cells: \[ E_{cell} = E^\circ_{cell} - \frac{0.06}{n}\log Q \] At \(298\,K\), \[ \frac{2.303RT}{F}=0.06V. \] Always determine the overall cell reaction first, then write the reaction quotient \(Q\).

Answer 1

The Correct Option is A

Step 1: Find the standard cell potential. The hydrogen electrode has higher reduction potential and acts as cathode. \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] \[ = 0-(-0.14) \] \[ = 0.14V \]

Step 2: Write the overall cell reaction. Anode: \[ M \rightarrow M^{2+}+2e^- \] Cathode: \[ 2H^+ +2e^- \rightarrow H_2 \] Overall reaction: \[ M+2H^+ \rightarrow M^{2+}+H_2 \] Thus, \[ n=2 \] and \[ Q = \frac{[M^{2+}]\,P_{H_2}} {[H^+]^2} \] Since \[ P_{H_2}=1, \] \[ Q=\frac{x}{(0.02)^2} =\frac{x}{4\times10^{-4}} =2500x. \]

Step 3: Apply Nernst equation. \[ E_{cell} = E^\circ_{cell} -\frac{0.06}{2}\log Q \] \[ 0.077 = 0.14 -0.03\log(2500x). \] \[ 0.03\log(2500x) = 0.14-0.077 = 0.063. \] \[ \log(2500x) = \frac{0.063}{0.03} = 2.1. \] \[ 2500x = 10^{2.1}. \] \[ x = \frac{10^{2.1}}{2500}. \] Now, \[ 2500 = 25\times100 \] \[ \log 2500 = \log25+2 = 2(0.699)+2 = 3.398. \] Hence, \[ \log x = 2.1-3.398 = -1.298. \] \[ x = 10^{-1.298} \approx 0.05. \]

Final Answer: \[ \boxed{x=0.05} \] \[ \boxed{\text{Answer = (A)}} \]