<page_title> 1840 United States presidential election in Rhode Island </page_title> <section_title> Results </section_title> <table> <cell> William Henry Harrison of Ohio <col_header> United States presidential election in Rhode Island, 1840 </col_header> <col_header> Candidate </col_header> <col_header> Count </col_header> </cell> <cell> 61.22% <col_header> United States presidential election in Rhode Island, 1840 </col_header> <col_header> Popular vote </col_header> </cell> </table>
In 1840 United States presidential election in Rhode Island, Harrison received 61.22% of the popular vote.