1912 United States presidential election in Rhode Island 1912 United States presidential election in Rhode Island The 1912 United States presidential election in Rhode Island took place on November 5, 1912, as part of the 1912 United States Presidential Election which was held throughout all contemporary 48 states. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president. Rhode Island was won by the Republican nominees, New Jersey Governor Woodrow Wilson and Indiana Governor Thomas R. Marshall. Wilson and Marshall defeated incumbent President William Howard Taft, and his running mate Vice President James S. Sherman and Progressive Party candidates, former President
1912 United States presidential election as president, Roosevelt challenged Taft at the 1912 Republican National Convention. After Taft and his conservative allies narrowly prevailed at the Republican convention, Roosevelt rallied his progressive supporters and launched a third party bid. Backed by William Jennings Bryan and other progressives, Wilson won the Democratic Party's presidential nomination on the 46th ballot, defeating Speaker of the House Champ Clark and several other candidates. Meanwhile, the Socialist Party renominated its perennial standard-bearer, Eugene V. Debs. The election of 1912 was bitterly contested by three individuals, Wilson, Roosevelt, and Taft, who all had or would serve as president. Roosevelt's "New Nationalism"