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Calculate Delta ,G^@ and log Kc for the...

Calculate `Delta ,G^@` and log `K_c` for the following reation:
`Cd^(+2)(aq)+Zn^(2+)(aq)+Cd(s)`
Given : `E_(cd^(2+)//cd)^(0)=0.403 V`
`E_(Zn^2+//Zn)^(0)=0.763 V`

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To solve the problem, we need to calculate \( \Delta G^\circ \) and \( \log K_c \) for the given reaction: \[ \text{Cd}^{2+}(aq) + \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + \text{Cd}(s) \] ### Step 1: Identify the half-reactions From the given standard reduction potentials: ...
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Calculate the equilibrium constant for the reaction: Cd^(2+) (aq) +Zn(s) hArr Zn^(2+) (aq) +Cd (s) If E_(Cd^(2+)//Cd)^(Theta) = -0.403V E_(Zn^(2+)//Zn)^(Theta) = -0.763 V

Calculate the equilibrium constant for the reaction at 298 K Zn(s)+Cu^(2+)(aq)harr Zn^(2+)(aq)+Cu(s) Given " " E_(Zn^(2+)//Zn)^(@)=-0.76 V and E_(Cu^(2+)//Cu)^(@)=+0.34 V

Calculate Delta_(r)G^(@) and log K_(c) for the following reaction at 298 K. 2Cr_((s))+Cd_((aq))^(3+)+33Cd_((s)){Given :E^(@)""_(Cell")=+0.34V,IF=96500Cmol^(-1)]

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Write Nernst equation for the reaction: (i) 2Cr(s) +3Cd^(2+)(aq) to 2Cr^(3+) (aq) +3Cd(s)

Under standered condition Delta G^(@) for the reaction 2Cr(s)= 3Cd^(2+)(aq) rarr 2Cr_((a a))^(3+) +3Cd(s) is (E_(Cr^(3+)//Cr)^(@) = - 0.74 V, E_(Cd^(2+)//Cd)^(@) = - 0.4 V)

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