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