The standard e.m.f of the cell is given by:
\[
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.33 \, \text{V} - 0.77 \, \text{V} = 0.56 \, \text{V}
\]
Current flows from anode to cathode, and the anode is where oxidation occurs, so electrode A (Cr) will act as the anode, and electrode B (Fe) will act as the cathode. Therefore, the given statements are true, making option (4) correct.