Given the standard electrode potentials,K^{+} | vedclass

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(A) The reducing power of a metal is inversely proportional to its standard reduction potential. $A$ more negative reduction potential indicates a str

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$Ag < Hg < Cr < Mg < K$

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(A) The reducing power of a metal is inversely proportional to its standard reduction potential. $A$ more negative reduction potential indicates a stronger reducing agent.The given standard reduction potentials are:$K^{+}/K = -2.93 \, V$$Mg^{2+}/Mg = -2.37 \, V$$Cr^{3+}/Cr = -0.74 \, V$$Hg^{2+}/Hg = 0.79 \, V$$Ag^{+}/Ag = 0.80 \, V$Arranging these in increasing order of reduction potential: $K Since the reducing power increases as the reduction potential becomes more negative,the increasing order of reducing power is: $Ag < Hg < Cr < Mg < K$.

List-$I$	List-$II$
$(A)$ Potential of hydrogen electrode at $pH = 10$	$(I)$ $0.76 \ V$
$(B)$ $Cu^{2+} \| Cu$	$(II)$ $0.059$
$(C)$ $Zn \| Zn^{2+}$	$(III)$ $-0.591 \ V$
$(D)$ $\frac{2.303 RT}{F}$	$(IV)$ $0.337 \ V$
	$(V)$ $-0.76 \ V$

Given the standard electrode potentials,$K^{+}/K = -2.93 \, V$,$Ag^{+}/Ag = 0.80 \, V$,$Hg^{2+}/Hg = 0.79 \, V$,$Mg^{2+}/Mg = -2.37 \, V$,and $Cr^{3+}/Cr = -0.74 \, V$,arrange these metals in their increasing order of reducing power.

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