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3 years ago

Calculate the electrical energy per gram of anode material for the following reaction at 298 K:

2 answers:

8 0

The substance oxidized is MnO2 which means that the oxidizing agent and at the same time the anode material is Li. The standard cell potential is already given and it is given in terms of per mole of reactant. So, the electrical energy per gram of anode is
3.15 / 7 = 0.45 V / g of Li

antiseptic1488 [7]3 years ago

6 0

The answer is:

E per gram = 0.45 V

The explanation:

when MnO2 is the substance who oxidized here so, the oxidizing agent and the anode here is Li.

and when the molar mass of Li is = 7 g/mol

and in our reaction equation we have 1 mole of Li will give 3.15 V of the electrical energy

that means that :

7 g of Li gives → 3.15 V

So 1 g of Li will give→ ???

∴ The E per gram = 3.15 V / 7 g of Li

= 0.45 V

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vesna_86 [32]

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The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

$Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}$

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The expression for solubility constant for this reaction will be:

$K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2$

We are given:

$K_{sp}=2.5\times 10^{-33}$

Putting values in above equation, we get:

$2.5\times 10^{-33}=(3s)^3\times (2s)^2\\\\2.5\times 10^{-33}=108s^5\\\\s=1.18\times 10^{-7}mol/L$

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