Answer : The standard cell potential of the reaction is, -1.46 V
Explanation :
The given balanced cell reaction is,
Here, chromium (Cr) undergoes oxidation by loss of electrons and act as an anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
The standard values of cell potentials are:
Standard reduction potential of lead ![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)
Standard reduction potential of chromium ![E^0_{[Cr^{3+}/Cr]}=1.33V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D%3D1.33V)
Now we have to calculate the standard cell potential for the following reaction.
![E^0=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cr^{3+}/Cr]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D)
Therefore, the standard cell potential of the reaction is, -1.46 V
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They determined that (1) An atom is mostly empty space with a sense, positively charged nucleus, because most particles passed straight through the gold foil, but occasionally they would deflect, sometimes by a very large margin
Answer:
Option E)
Explanation:
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<u>1. Write the solubility equation:</u>
- CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻
<u>2. Write the concentrations below the equation:</u>
- CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻
A - s s s
<u>3. Write the Ksp equation:</u>
- Ksp = [Ca²⁺] . [CO₃²⁻] ↔ the solid substances do not appear
<u>4. Solve the equation:</u>
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