TS EAMCET · Chemistry · Electrochemistry

The $EMF$ of a galvanic cell consisting of two hydrogen electrodes is $0.17 V$ . If the solution of one of the electrodes has $[H^{+}] = 10^{- 3} M$ , the $pH$ at the other electrode is:

A $5.88$
B $4.88$
C $2.08$
D $3.08$

Verified Solution

Answer & Solution

Correct Answer

(A) $5.88$

Step-by-step Solution

Detailed explanation

Ecell=-0.059 logH+electrode 1H+electrode 2 0.17 V=-0.059 logH+10-3 -2.88=logH+-log10-3 -2.88=logH++3 -logH+=3+2.88=5.88 pH=5.88

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The EMF of a galvanic cell consisting of two hydrogen electrodes is 0.17 V. If the solution of one of the electrodes has H+=10-3 M, the pH at the other electrode is:

Step-by-step Solution

The $EMF$ of a galvanic cell consisting of two hydrogen electrodes is $0.17 V$ . If the solution of one of the electrodes has $[H^{+}] = 10^{- 3} M$ , the $pH$ at the other electrode is: