Answer : The correct option is, +91 kJ/mole
Solution :
The balanced cell reaction will be,
Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)
![E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}](https://tex.z-dn.net/?f=E%5E0_%7Bcell%7D%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D)
Now we have to calculate the standard Gibbs free energy.
Formula used :
where,
= standard Gibbs free energy = ?
n = number of electrons = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = -0.47 V
Now put all the given values in this formula, we get the Gibbs free energy.
Therefore, the standard Gibbs free energy is +91 kJ/mole
Answer : If we list the given chemicals according to their increasing oxidising ability then the order will be like this; 1 being the strongest and 6 being the weakest
1. K > 2. Ca >3. Ni> 4. Cu> 5. Ag> 6.Au
Explanation : Considering the reduction potential of each chemical species it will be easy to identify their oxidising capacity and differentiate accordingly;
More negative the value of reduction potential more is the ability of the chemical species to get oxidised.
Chemicals with their reduction potential is given below.
K has -2.92; Ca has -2.76; Ni has -0.23; Cu has 0.52; Ag has 1.50 and Au has 1.50.
False that atom is the smallest identifiable unit of a compound.
The smallest identifiable unit of a compound is the Element. Element is the one which make up the compound and element is made up by atoms. Example of element is oxygen and hydrogen which make up water (H2O) which is a compound.
Speed in km/hr = 15 x 18
------------
5
= 54 km/hr.
Hope this helps!
