Consider the following reaction

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{+2}+4H_{2}O,E°=1.51V.$

The quantity of electricity required is Faraday to reduce five moles of $Mn{O}_4^{-}$  is--------.

(Integer answer)

<p><math><mrow><mi>M</mi><mi>n</mi><msubsup><mrow><mi>O</mi></mrow><mrow><mn>4</mn></mrow><mrow><mo>−</mo></mrow></msubsup><mo>+</mo><mn>8</mn><mi>H</mi><mo>+</mo><mn>5</mn><msup><mrow><mi>e</mi></mrow><mrow><mo>−</mo></mrow></msup><mo>→</mo><mi>M</mi><mi>n</mi><mo>+</mo><mn>4</mn><msub><mrow><mi>H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi>O</mi><mtext> </mtext><mtext> </mtext><msup><mrow><mi>E</mi></mrow><mrow><mi>o</mi></mrow></msup><mo>=</mo><mn>1</mn><mo>.</mo><mn>5</mn><mn>1</mn><mi>V</mi></mrow></math></p><p>Quantity of electricity required to reduce 1 mole of&nbsp;<math><mrow><mi>M</mi><mi>n</mi><msubsup><mrow><mi>O</mi></mrow><mrow><mn>4</mn></mrow><mrow><mo>−</mo></mrow></msubsup></mrow></math>&nbsp;is 5F</p><p>So, for 5 mole&nbsp;<math><mrow><mi>M</mi><mi>n</mi><msubsup><mrow><mi>O</mi></mrow><mrow><mn>4</mn></mrow><mrow><mo>−</mo></mrow></msubsup></mrow></math>&nbsp;25F electricity is required.</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment