ⓘ 1860 United States presidential election in Rhode Island

                                     

ⓘ 1860 United States presidential election in Rhode Island

The 1860 United States presidential election in Rhode Island took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose 4 electors of the Electoral College, who voted for president and vice president.

Rhode island won the Republican partys candidate Abraham Lincoln, who won by a margin of 22.74%.

With 61.37% of the vote, Rhode island would be the Lincolns fifth strong state from the point of view is interesting percentage of the vote in the elections of 1860, after Vermont, Minnesota, Massachusetts and Maine.

Like new Jersey, new York and Pennsylvania, Rhode island, was one of four States that a fusion ticket of Democrats, which included not only Northern Democrats, but supporters of the southern Democrats and the constitutional leaders, as well.

                                     
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