Given that E° values of Ag+/Ag, K+/K, Mg2+/Mg and Cr3+/Cr are 0.80 V, -2.93 V, -2.37 V and -0.74 V respectively. Therefore the order for the reducing power of the metal is

Question

Given that E° values of Ag+/Ag, K+/K, Mg2+/Mg and Cr3+/Cr are 0.80 V, -2.93 V, -2.37 V and -0.74 V respectively. Therefore the order for the reducing power of the metal is (A) Ag > Cr > Mg > K (B) Ag < Cr < Mg < K (C) Ag > Cr > K > Mg (D) Cr > Ag > Mg > K

Tardigrade Admin · Accepted Answer

More the negative E° value, larger the reducing power of the metal The greater the standard reduction potential, the greater tendency to reducing power of metal. K has greater value (-2.93) so K > Mg > Cr > Ag

Q. Given that $E °$ values of $A g^{+} / A g, K^{+} / K, M g^{2 +} / M g$ and $C r^{3 +} / C r$ are $0.80 V, - 2.93 V, - 2.37 V$ and $- 0.74 V$ respectively. Therefore the order for the reducing power of the metal is

$A g > C r > M g > K$

$A g < C r < M g < K$

$A g > C r > K > M g$

$C r > A g > M g > K$

More the negative $E^{°}$ value, larger the reducing power of the metal The greater the standard reduction potential, the greater tendency to reducing power of metal. $K$ has greater value $(- 2.93)$ so $K > M g > C r > A g$

Q. Given that E° values of Ag+/Ag,K+/K,Mg2+/Mg and Cr3+/Cr are 0.80V,−2.93V,−2.37V and −0.74V respectively. Therefore the order for the reducing power of the metal is

Ag>Cr>Mg>K

Ag<Cr<Mg<K

Ag>Cr>K>Mg

Cr>Ag>Mg>K