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Electrochemistry question

2022 · Q99
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Electrochemistry question

2022 · Q99

NEETChemistryElectrochemistryMCQ+4 / −1

Two half cell reactions are given below.

Co3++e−→Co2+,         ECo2+/Co3+0=−1.81 VC{o^{3 + }} + {e^ - } \to C{o^{2 + }},\,\,\,\,\,\,\,\,\,E_{C{o^{2 + }}/C{o^{3 + }}}^0 = - 1.81\,VCo3++e−→Co2+,ECo2+/Co3+0​=−1.81V

2Al3++6e−→2Al(s),   EAl/Al3+0=+1.66 V2A{l^{3 + }} + 6{e^ - } \to 2Al(s),\,\,\,E_{Al/A{l^{3 + }}}^0 = + 1.66\,V2Al3++6e−→2Al(s),EAl/Al3+0​=+1.66V

The standard EMF of a cell with feasible redox reaction will be :

  1. A
    −-−3.47 V
  2. B
    +7.09 V
  3. C
    +0.15 V
  4. D
    +3.47 V
View written solutionFree

Correct answer: D

Since $E_{OP}^0$ of Al is more than Co2+, so at anode Al will oxidise and at cathode Co3+ will reduce.

$$E_{Cell}^0 = {(E_{Cathode}^0)_{RP}} - {(E_{Anode}^0)_{RP}}$$

$$ = E_{C{o^{3 + }}/C{o^{2 + }}}^0 - E_{A{l^{3 + }}/Al}^0$$

$ = (1.81) - ( - 1.66)$

$ = + 3.47\,V$

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