Question

Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Fe+ (aq) + Cd(s)— 2Fe2+ (aq) + Cd²+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant AGⓇ for this reaction would be than zero. Submit Answer Retry Entire Group 6 more group attempts remaining

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Questions: in....typed format...

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$\Rightarrow$Question - 1:

$\Rightarrow$Answer:

1. Equilibrium constant:  K_eq = 3.3653 x 10³⁹   
2. The value of standard: Gibb's free energy: of the reaction:  $\Delta$G^o  , will be less than zero , i.e. negative...

$\Rightarrow$Explanation:

• Given:

1. ​​​​​​​ Cell Reaction:  2 Fe³⁺ (aq) + Cd (s)  $\rightarrow$ 2 Fe²⁺ (aq) + Cd²⁺ (aq)

• ​​​​​​​Step - 1:

​​​​​​​We can use the following half cell reactions: and standard potentials: for Oxidation and Reduction: adding up which, will give us the given, balanced cell reaction: and the standard cell potential ( E^o_cell  )

Type of Reaction	Half Cell Reaction	standard electrode potential
Reduction ( cathode )	2 Fe3+ (aq) + 2e-  $\rightarrow$ 2 Fe2+ (aq)	Eored = + 0.77 V ( volts )
Oxidation ( anode )	Cd (s)  $\rightarrow$ Cd2+ (aq) + 2e-	Eoox = + 0.40 V ( volts )
Adding above equations: Final	2 Fe3+ (aq) + Cd (s)  $\rightarrow$ 2 Fe2+ (aq) + Cd2+ (aq)	Eocell = ( 0.77 + 0.40 ) = + 1.17 V ( volts )

$\Rightarrow$We get: standard cell potential: E^o_cell = + 1.17 V ( volts )

$\Rightarrow$ Number of electrons, exchanged: n = 2

• Step - 2:

​​​​​​​We know, the following formula:

$\Rightarrow$   E^o_cell = (  0.0592 / n ) x  log₁₀ K_eq-------------------------------Equation - 1

Plugging in values in Equation - 1, above, we will get the following:

$\Rightarrow$    +1.17 V = (  0.0592 / 2 ) x  log₁₀ K_eq

$\Rightarrow$   log₁₀ K_eq = 39.527

$\Rightarrow$   K_eq  = 3.3653 x 10³⁹   

• Step - 3:

​​​​​​​We know:   $\Delta$G^o =  - RT x ln ( K_eq ) = - ( 8.314 J / mol. K ) x ( 298.15 K ) x ln ( 3.3653 x 10³⁹ ) = - 225608.07 J / mol

$\Rightarrow$ i.e.   $\Delta$G^o = - 225.608 kJ  ( kilo-joule )

$\Rightarrow$ Therefore: the value of :  $\Delta$G^o  , will be less than zero , i.e. negative...

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$\Rightarrow$Question - 2:

$\Rightarrow$Answer:

1. Equilibrium constant:  K_eq = 8.2985 x 10^-7  
2. The value of standard: Gibb's free energy: of the reaction:  $\Delta$G^o  , will be greater than zero , i.e. POSITIVE...

​​​​​​​$\Rightarrow$Explanation:

• Given:

1. ​​​​​​​ Cell Reaction:  2 Cu²⁺(aq) + Cu (s)  $\rightarrow$ 2 Cu⁺ (aq) + Cu²⁺ (aq)

• ​​​​​​​Step - 1:

​​​​​​​We can use the following half cell reactions: and standard potentials: for Oxidation and Reduction: adding up which, will give us the given, balanced cell reaction: and the standard cell potential ( E^o_cell  )

Type of Reaction	Half Cell Reaction	standard electrode potential
Reduction ( cathode )	2 Cu2+ (aq) + 2e-  $\rightarrow$ 2 Cu+ (aq)	Eored = + 0.16 V ( volts )
Oxidation ( anode )	Cu(s)  $\rightarrow$ Cu2+ (aq) + 2e-	Eoox = - 0.34 V ( volts )
Adding above equations: Final	2 Cu2+(aq) + Cu (s)  $\rightarrow$ 2 Cu+ (aq) + Cu2+ (aq)	Eocell = ( + 0.16 - 0.34 ) = - 0.18 V ( volts )

$\Rightarrow$We get: standard cell potential: E^o_cell = - 0.18 V ( volts )

$\Rightarrow$ Number of electrons, exchanged: n = 2

• Step - 2:

​​​​​​​We know, the following formula:

$\Rightarrow$   E^o_cell = (  0.0592 / n ) x  log₁₀ K_eq-------------------------------Equation - 1

Plugging in values in Equation - 1, above, we will get the following:

$\Rightarrow$    - 0.18 V = (  0.0592 / 2 ) x  log₁₀ K_eq

$\Rightarrow$   log₁₀ K_eq = - 6.081

$\Rightarrow$   K_eq  = 8.2985 x 10^-7  

• Step - 3:

​​​​​​​We know:   $\Delta$G^o =  - RT x ln ( K_eq ) = - ( 8.314 J / mol. K ) x ( 298.15 K ) x ln ( 8.2985 x 10^-7​​​​​​​   ) = -34708.476 J / mol

$\Rightarrow$ i.e.   $\Delta$G^o = 34.708  kJ  ( kilo-joule )

$\Rightarrow$ Therefore: the value of :  $\Delta$G^o  , will be greater than zero , i.e. POSITIVE...

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