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Electrochemistry question

2023 · 25 Jan · Shift 2 · Q17
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  5. /2023 · 25 Jan · Shift 2 · Q17

Electrochemistry question

2023 · 25 Jan · Shift 2 · Q17

JEE MainChemistryElectrochemistryNumerical+4 / −1
Pt(s)∣H2(g)(1 bar)∣H+(aq)(1 M)∣∣M3+(aq),M+(aq)∣Pt(s)Pt(s)|{H_2}(g)(1\,bar)|{H^ + }(aq)(1\,M)||{M^{3 + }}(aq),{M^ + }(aq)|Pt(s)Pt(s)∣H2​(g)(1bar)∣H+(aq)(1M)∣∣M3+(aq),M+(aq)∣Pt(s) The Ecell\mathrm{E_{cell}}Ecell​ for the given cell is 0.1115 V at 298 K when [M+(aq)][M3+(aq)]=10a{{\left[ {{M^ + }(aq)} \right]} \over {\left[ {{M^{3 + }}(aq)} \right]}} = {10^a}[M3+(aq)][M+(aq)]​=10a The value of aaa is ‾\underline{\hspace{2cm}}​ Given : EM3+/M+θ=0.2\mathrm{E_{{M^{3 + }}/{M^ + }}^\theta = 0.2}EM3+/M+θ​=0.2 V 2.303RTF=0.059V{{2.303RT} \over F} = 0.059VF2.303RT​=0.059V
Numerical answer
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