Consider the following relations for EMF of a | vedclass

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(D) The $EMF$ of a cell is defined as:$EMF_{cell} = E_{cathode} (reduction) - E_{anode} (reduction)$Since $E_{anode} (reduction) = -E_{anode} (oxidati

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$(ii)$ and $(iv)$

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(D) The $EMF$ of a cell is defined as:$EMF_{cell} = E_{cathode} (reduction) - E_{anode} (reduction)$Since $E_{anode} (reduction) = -E_{anode} (oxidation)$,we can write:$EMF_{cell} = E_{cathode} (reduction) + E_{anode} (oxidation)$Also,since $E_{cathode} (reduction) = -E_{cathode} (oxidation)$,we can write:$EMF_{cell} = E_{anode} (oxidation) - E_{cathode} (oxidation)$Comparing these with the given relations:$(ii)$ $EMF$ of cell = (Oxidation potential of anode) $+$ (Reduction potential of cathode) is correct.$(iv)$ $EMF$ of cell = (Oxidation potential of anode) $-$ (Oxidation potential of cathode) is correct.Therefore,the correct option is $D$.

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