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The half-cell reduction potential of a hudrogen electrode at `pH= 10` will be.A. ` 0. 95 V`B. ` -0. 59 V`C. ` 0.059 V`D. ` -0 .059 ` |
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Answer» Correct Answer - B `E_(red = E_(cell) ^@ + (0. 0591)/n log [Mn^(=)` `= 0 + (0.0591)/1 log [10^(-10)]` ` =+ 0.0591 xx - 10 - 0. 59 V`. |
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