Ncert Exercises & Solutions

6.23 The choice of a reducing agent in a particular case depends on thermodynamics factor. How far do you agree with this statement? Support your opinion with two examples.

The above figure is a plot of Gibbs energy

(∆ G^{θ})

vs. T for formation of some oxides. It can be observed from the above graph that metal can reduce the oxide of other metals, if the standard free energy of formation

(∆_{t} G^{θ})

of the oxide of the former is more negative than the latter. For example, since

∆_{t} {G^{θ}}_{(A I, A I_{2}, O_{3})}

is more negative than ,

∆_{t} {G^{θ}}_{(C u, C u_{2}, O)}

Al can reduce Cu2O to Cu, but Cu cannot reduce

A I_{2} O_{3}

. Similarly, Mg can reduce ZnO to Zn, but Zn cannot reduce MgO because

∆_{t} {G^{θ}}_{(M g, M g O)}

is more negative than

∆_{t} {G^{θ}}_{(Z n, Z n O)}

In the electrolysis of molten NaCl,

C L_{2}

is obtained at the anode as a by product.

N a C L_{(m e l t)} \to N {a^{+}}_{(m e l t)} + C {L^{-}}_{(m e l t)}

A t c a t h o d e : N {a^{+}}_{(m e l t)} + e^{-} \to N a_{(s)}

A t a n o d e : C {l^{-}}_{(m e l t)} \to C l_{(g)} + e^{-}

2 C L_{(g)} \to C l_{2 (g)}

T h e o v e r a l l r e a c t i o n i s a s f o l l o w s :

N a C L_{(m e l t)} \overset{E l e c t r o l y s i s}{\to} N a_{(s)} + \frac{1}{2} C l_{2 (g)}

If an aqueous solution of NaCl is electrolyzed,

C l_{2}

will be obtained at the anode but at the cathode,

H_{2}

will be obtained (instead of Na). This is because the standard reduction potential of Na (E°= − 2.71 V) is more negative than that of

H_{2} O

(E° = − 0.83 V). Hence,

H_{2} O

will get preference to get reduced at the cathode and as a result,

H_{2}

is evolved.

N a C L_{(a q)} \to N {a^{+}}_{(a q)} + C {L^{-}}_{(a q)}

A t c a t h o d e : 2 H_{2} O_{(l)} + 2 e^{-} \to H_{2 (g)} + 2 O H_{(a q)}^{-}

A t a n o d e : C {l^{-}}_{(m e l t)} \to C l_{(g)} + e^{-}

2 C L_{(g)} \to C l_{2 (g)}

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