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Consider the following cell If Ka of HA and HB are 10⁻⁶ and 10^ -x . If e.m.f of cell at 298 K is 0.09 V and 2.303 RT F =0.06 . The value of x is

Correct answer

3

Step-by-step solution

Cathode :- 2 HB +2 e ⁻ H ₂ Anode :- 2 HA H ₂ aligned & E_ cell =E_ cell ^o- 0.06 2 10⁻⁶ 10^ -x & 0.09=-0.03 10^ -6+x & =-6+x & -3=-6+x & x=3 aligned

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