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Electrochemistry question

2023 · Q96
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Electrochemistry question

2023 · Q96

NEETChemistryElectrochemistryMCQ+4 / −1

The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is

Fe(aq)2++2e−→Fe(s),E∘=−0.44 VCr2O72− (aq) +14H++6e−→2Cr3++7H2OE∘=+1.33 V\begin{aligned} & \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}), \mathrm{E}^{\circ}=-0.44 \mathrm{~V} \\ & \mathrm{Cr}_2 \mathrm{O}_7^{2-} \text { (aq) }+14 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{E}^{\circ}=+1.33 \mathrm{~V} \end{aligned}​Fe(aq)2+​+2e−→Fe(s),E∘=−0.44 VCr2​O72−​ (aq) +14H++6e−→2Cr3++7H2​OE∘=+1.33 V​

  1. A
    +1.77 V+1.77 \mathrm{~V}+1.77 V
  2. B
    +2.65 V+2.65 \mathrm{~V}+2.65 V
  3. C
    +0.01 V+0.01 \mathrm{~V}+0.01 V
  4. D
    +0.89 V+0.89 \mathrm{~V}+0.89 V
View written solutionFree

Correct answer: A

$$\begin{aligned} \mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\mathrm{C}}^{\mathrm{o}}-\mathrm{E}_{\mathrm{A}}^{\mathrm{o}} \\ & =(1.33)-(-0.44) \\ & =+1.77 \mathrm{~V} \end{aligned}$$

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