Question 30 of 31intermediate🔧 ApplyShort Answer2 marks

Given the standard electrode potentials, K⁺/K = -2.93V, Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V, Mg²⁺/Mg = -2.37V, Cr³⁺/Cr = -0.74V arrange these metals in their increasing order of reducing power.

Correct Answer

Ag < Hg < Cr < Mg < K

Exercise: EXERCISES | Q: 7.29 | (Page Page 21)
For More Understanding

Explanation

The reducing power of a metal is inversely proportional to its standard electrode potential. A lower (more negative) standard electrode potential indicates a higher tendency to lose electrons, making the metal a stronger reducing agent. The given potentials are Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V, Cr³⁺/Cr = -0.74V, Mg²⁺/Mg = -2.37V, and K⁺/K = -2.93V. Arranging these in decreasing order of their electrode potentials gives the increasing order of their reducing power.

Solution Steps

  1. Step 1: Understand the relationship between standard electrode potential and reducing power. A lower reduction potential means a greater capability to lose electrons, hence a stronger reducing agent.

  2. Step 2: List the given metals with their standard electrode potentials: Ag = 0.800.80V, Hg = 0.790.79V, Cr = −0.74-0.74V, Mg = −2.37-2.37V, K = −2.93-2.93V.

  3. Step 3: Arrange the metals in decreasing order of their electrode potentials: Ag (0.800.80V) > Hg (0.790.79V) > Cr (−0.74-0.74V) > Mg (−2.37-2.37V) > K (−2.93-2.93V).

  4. Step 4: Reverse this order to get the increasing order of reducing power: Ag < Hg < Cr < Mg < K.

PreviousNext