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Consider the following cell reaction : 2 AgCl ( ~s )+ H ₂( ~g ) Ag ( ~s )+2 H ⁺( aq )+2 Cl ⁻( aq ) E ^ =0.19 ~V At [ Cl ⁻ ]=10⁻³ M and P ( H ₂ )=0.01 ~atm , find the pH , if the cell potential is 0.43 V at 25^ C . (Take 2.303 R(298) F =0.06 )

Correct answer

2

Step-by-step solution

aligned & E = E ^ - 0.06 2 [ H ⁺ ]^2 [ Cl ⁻ ]^2 P _ H ₂ & 0.43=0.19- 0.06 2 [ H ⁺ ]^2 (10⁻³ )^2 10⁻² & 0.43=0.19- 0.06 2 [ H ⁺ ]^2 (10⁻² )^2 & 0.43=0.19-0.06 ( H ⁺ )+0.12 & 0.12=0.06 pH & pH =2 aligned

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