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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 $D$.

Similar Questions

The standard reduction potentials for $Cu^{2+}|Cu$ and $Cu^{2+}|Cu^{+}$ are $0.337 \, V$ and $0.153 \, V$ respectively. What is the standard electrode potential for the $Cu^{+}|Cu$ half-cell in $V$?

Given the standard electrode potentials:
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Arrange these metals in their increasing order of reducing power.

The standard reduction potential for $Fe^{2+}/Fe$ and $Sn^{2+}/Sn$ electrodes are $-0.44 \ V$ and $-0.14 \ V$ respectively. For the cell reaction,$Fe^{2+} + Sn \longrightarrow Fe + Sn^{2+}$,the standard emf is:

Standard reduction potentials at $25^\circ C$ of $Li^{+}|Li$,$Ba^{2+}|Ba$,$Na^{+}|Na$,and $Mg^{2+}|Mg$ are $-3.05 \ V$,$-2.90 \ V$,$-2.71 \ V$,and $-2.37 \ V$ respectively. Which one of the following is the strongest oxidising agent?

The standard reduction potentials of $4$ elements are given below. Which of the following will be the most suitable reducing agent?
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