Arrange the following redox couples in the increasing order of their reducing strength: [A]=Cu∕Cu2+‌E0=−0.34V [B]=Ag∕Ag+‌E0=−0.8V [C]=Ca∕Ca2+‌E0=+2.87V [D]=Cr∕Cr3+‌E0=+0.74V
To determine the increasing order of the reducing strength of the redox couples, we need to analyze their standard electrode potentials ( E0 values). Remember, the lower (more negative) the standard electrode potential, the stronger the reducing agent, because it has a stronger tendency to lose electrons. The given redox couples and their standard electrode potentials are: [A]=Cu∕Cu2+‌E0=−0.34V [B]=Ag∕Ag+‌E0=−0.8V [C]=Ca∕Ca2+‌E0=+2.87V [D]=Cr∕Cr3+‌E0=+0.74V Let's arrange these redox couples in the increasing order of their reducing strength based on their E0 values: Ag∕Ag+has the lowest E0 value of -0.8 V , making it the strongest reducing agent. Cu∕Cu2+ has an E0 value of -0.34 V , making it the next strongest reducing agent. Cr∕Cr3+ has an E0 value of +0.74 V , making it a weaker reducing agent. Ca∕Ca2+ has the highest E0 value of +2.87 V , making it the weakest reducing agent. Therefore, the increasing order of reducing strength is: B<A<D<C So, the correct answer is: Option A: B < A < D < C